Question: Multiply the following complex numbers: $({-4-4i}) \cdot ({2+4i})$
Explanation: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4-4i}) \cdot ({2+4i}) = $ $ ({-4} \cdot {2}) + ({-4} \cdot {4}i) + ({-4}i \cdot {2}) + ({-4}i \cdot {4}i) $ Then simplify the terms: $ (-8) + (-16i) + (-8i) + (-16 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -8 + (-16 - 8)i - 16i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -8 + (-16 - 8)i - (-16) $ The result is simplified: $ (-8 + 16) + (-24i) = 8-24i $